Sequence $ (a_n)$ is called geometric sequence if every member starting from the second is equal to the first member multiplied by some constant $ q, q \not= 0$.
From the definition we can conclude that a particular sequence is geometric if quotient of every member of the sequence (except first) and the member in front of it is a constant and equal to $q$.
$ q = \frac{a_{n + 1}}{a_n}$
Number $q$ is called quotient of geometric sequence.
Geometric sequence is uniquely identified by its first member and quotient.
Example 1. Continue the sequence: $2, 8, 32$ until you find three new members.
Now that we know few kinds of sequences we have to check which sequence this is. Since the difference between the first two members is $6$, and the difference between second and third member is $24$ this can’t be arithmetic sequence. Now we can check for quotient. $ 8 : 2 = 4$ and $ 32 : 8 = 4$. From this it is safe to conclude that this is geometric sequence with quotient $4$.
Now, we just continue multiplying with $4$ and we get this sequence:
$2, 8, 32, 128, 512, 2048, …$
Geometric sequence got its name by the link between geometric middle and its members. Every member of geometric sequence is equal to geometric middle of its adjacent members.
$ a_n = \sqrt{a_{n + 1} \cdot a_{n – 1}}$
Example 2. Check if the previous statement is true on the sequence:
$3, 6, 12, 24, 48$.
$ a_2 = \sqrt{3 \cdot 12} = \sqrt{36} = 6$
$ a_3 = \sqrt{6 \cdot 24} = \sqrt{144} = 12$
$ a_4 = \sqrt{48 \cdot 12} = \sqrt{576} = 24$
If a geometric sequence has quotient equal to $1$, then that sequence is a constant sequence.
General member of geometric sequence
Just like in arithmetic sequence we’ll start from the definition of geometric sequence in order to find general member of geometric sequence.
$ a_2 = a_1 \cdot q$
$ a_3 = a_2 \cdot q = a_1 \cdot q^2$
$ a_4 = a_3 \cdot q = a_2 \cdot q^2 = a_1 \cdot q^3$
…
$ a_n = a_1 \cdot q_{n – 1}$
General member of geometric sequence with first member $ a_1$ and quotient $q \not= 0$ has a form $ a_n = a_1 \cdot q_{n – 1}$
Example 3. Find seventh member of a sequence $\frac{1}{2^n}$.
$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}$
First member is $\frac{1}{2}$, and the quotient is $\frac{1}{2}$.
$ a_7 = a_1 \cdot q_6$
$ a_7 = \frac{1}{2} \cdot (\frac{1}{2})^6$
$ a_7 = \frac{1}{2} \cdot \frac{1}{64}$
$ a_7 = \frac{1}{128}$
Example 4. Find geometric sequence whose second member is $24$ and fifth $81$.
If we want to find geometric sequence we must find the first member and the quotient.
We’ll use the definition of a general member of geometric sequence.
$ a_2 = a_1\cdot q \Rightarrow 24 = a_1 \cdot q$
$ a_5 = a_1 \cdot q^4 \Rightarrow 81 = a_1 \cdot q^4$
The easiest way to solve this system is to divide those two equations.
$ q^3 = \frac{27}{8}$
$ q = \frac{3}{2}$
$ a_1 = 16$
The sum of the first n members of geometric sequence.
$ S_n = a_1 + a_2 + a_3 +…+ a_n$
$ S_n = a_1 + a_1 \cdot q + a_1 \cdot q^2 +…+ a_1\cdot q^{n – 1}$
By multiplying this expression with q and subtracting from $ S_n$ we get:
$ S_n = a_1 + a_1 \cdot q + a_1 \cdot q^2 +…+ a_1 \cdot q^{n – 1}$
$ q \cdot S_n = a_1 \cdot q + a_1 \cdot q^2 +…+ a_1 \cdot q^n$
$ S_n – q \cdot S_n = a_1 – a_1 \cdot q^n$
$ S_n (1 – q) = a_1 (1 – q^n)$
$ S_n = \frac{a_1 (1 – q^n)}{1 – q}$, $ q \not= 1$
If $ q = 1$, then this is a constant sequence $ a_1, a_2, a_3,…$ so $ S_n = n \cdot a_1$
Example 5. How many members of geometric sequence
$ 2, – 1, \frac{1}{2}, – \frac{1}{4}, \frac{1}{8}$
do you have to add so that the sum is equal to $\frac{85}{64}$?
This sequence is clearly not a constant sequence so we can calculate sum by the formula
$ S_n = \frac{a_1 (1 – q^n)}{1 – q}$
This kind of sequence whose members are changing signs in a right pattern (they are changing their sign with every following member) is called alternating sequence. In these kinds of sequences the quotient is negative.
$ a_1 = 2$
$ q = – \frac{1}{2}$
$ \frac{85}{64} = \frac{2 (1 – (-\frac{1}{2})^n)}{1 – (-\frac{1}{2})^n}$
$ \frac{85}{64} = \frac{2 (1 – (-\frac{1}{2})^n)}{\frac{3}{2}}$
$ \frac{255}{256} = 1 – (-\frac{1}{2})^n$
$ (-\frac{1}{2})^n = \frac{1}{256}$
$ (-\frac{1}{2})^n = (-\frac{1}{2})^8$
$ n = 8$
